Ball Dropped from Cliff

A stone is dropped from a cliff; 2.18 s later another stone is thrown downward with an initial speed of 31 m/s. They reach the ground simultaneously. Find the height of the cliff.

2. Relevant equations

vf^2 = vo^2 + 2ax
x = vot + 0.5at^2
vf = vo + at

3. The attempt at a solution

i've been working on this for the past two hours, and it's driving me nutsss.

i found the speed of the stone in 2.18 seconds by using the equation vf = vo + at. (t = 2.18 seconds; a = 9.81 m/s^2; vo = 0) i found the velocity to be 21.3858 m/s.

with that information, i found how far the stone traveled with the equation vf^2 = vo^2 + 2ax. (vf^2 = (21.3858)^2; vo^2 = 0; a = 9.81) the answer: 23.310522 m.

after that, i don't know what else to do with stone b. i've been playing around with the first equation, since the two stones hit the ground at the same time. is it wrong to assume that the final velocities of the two would be the same?

i would appreciate it very much if someone could throw me a bone on what to do with the second stone because i'm at a loss.

You have the right equations, and are plugging in values correctly. But you seem to be calculating unnececessary things.

Is it in any way useful to calculate how far the first stone has travelled after 2.18s?

Focus on thinking about what you really want, i.e. the height of the cliff. Think of x as being the height of the cliff. Then plug in values for your equations for the first stone, and then the second stone.

i'm still stuck on what i need to do.... i don't know where to go after finding the speed of the first stone at 2.18 seconds. i know a = 9.81 m/s^2, speed of the first stone is 21.3858 m/s when the second stone is thrown with an initial velocity of 31 m/s. after that, everything is coming up blank. i've tried to set equations equal to each other, but they always cancel out the wanted variable.